2.3.6 problem 46

Let $S$ be the event of successfully getting the Car under the specified strategy. Let $C_i$ be the event that Door $i$ contains the Car. Let $A$ be the event that Monty reveals the Apple, and let $A_i$ be the event that Door $i$ contains the Apple.

(a)

$$\begin{array}{cccc}P\left(S\right)& =P(S\cap C_1)+P(S\cap C_2)+P(S\cap C_3)+P(S\cap C_4)& & \\ & =P\left(C_1\right)P\left(S\right|C_1)+P(C_2\left)P\right(S\left|C_2\right)+P\left(C_3\right)P\left(S\right|C_3)+P(C_4\left)P\right(S\left|C_4\right)& & \\ & =\frac{1}{4}\ast 0+3\ast \frac{1}{4}\ast (p+q)\frac{1}{2}& & \\ & =3\ast \frac{1}{4}\ast \frac{1}{2}& & \\ & =\frac{3}{8}& & \end{array}$$

(b)

$$\begin{array}{cccc}P\left(A\right)& =P(A\cap G_1)+P(A\cap A_1)+P(A\cap B_1)+P(A\cap C_1)& & \\ & =P\left(G_1\right)P\left(A\right|G_1)+P(A_1\left)P\right(A\left|A_1\right)+P\left(B_1\right)P\left(A\right|B_1)+P(C_1\left)P\right(A\left|C_1\right)& & \\ & =\frac{1}{4}p+0+\frac{1}{4}q+\frac{1}{4}q& & \\ & =\frac{1}{4}(1+q)& & \end{array}$$

(c)

$$P\left(S\right|A)=\frac{P(S\cap A)}{P\left(A\right)}=\frac{\frac{1}{4}\ast p\ast \frac{1}{2}+\frac{1}{4}\ast q\ast \frac{1}{2}}{\frac{1}{4}(1+q)}=\frac{\frac{1}{8}}{\frac{1}{4}(1+q)}$$